Clifford traces and normalized spin completeness

Direct multiplication of the displayed Pauli matrices gives ${\sigma }_j^2=I_2$, ${\sigma }_1{\sigma }_2=i{\sigma }_3$, ${\sigma }_2{\sigma }_3=i{\sigma }_1$, ${\sigma }_3{\sigma }_1=i{\sigma }_2$, and the reversed products have the opposite signs. Hence ${\sigma }_i{\sigma }_j={\delta }_{ij}{I}_2+i{ϵ}_{ijk}{\sigma }_k$. Block multiplication then gives ${\alpha }_i^{†}={\alpha }_i$, ${\beta }^{†}=\beta$, ${\alpha }_i{\alpha }_j={\delta }_{ij}{I}_4+i{ϵ}_{ijk}{\mathrm{\Sigma }}_k$, and ${\alpha }_i\beta =-\beta {\alpha }_i$. The explicit blocks have trace zero, so $\mathrm{Tr}{\gamma }^a=0$, and $\mathrm{Tr}{I}_4=4$. The Clifford relation is $\{{\gamma }^a,{\gamma }^b\}=2h^{ab}{I}_4$. Cyclicity of trace follows by expanding matrix entries; averaging $\mathrm{Tr}({\gamma }^a{\gamma }^b)$ with its cyclic rearrangement gives $4h^{ab}$. For the four-factor trace, anticommute ${\gamma }^a$ successively past ${\gamma }^b,{\gamma }^c,{\gamma }^d$; cyclicity makes the final four-factor term the negative of the original, yielding $4(h^{ab}{h}^{cd}-h^{ac}{h}^{bd}+h^{ad}{h}^{bc})$. Finally ${\gamma }_5^2=I_4$ and ${\gamma }_5{\gamma }^a=-{\gamma }^a{\gamma }_5$ by the same Clifford relations, so cyclicity gives $\mathrm{Tr}({\gamma }^a{\gamma }^b{\gamma }^c)=-\mathrm{Tr}({\gamma }^a{\gamma }^b{\gamma }^c)$, hence the three-factor trace is zero.

The relations in 1.1 give $H_{\pm }^{†}=H_{\pm }$ and $H_{\pm }^2=(\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2+M^2)I_4=E^2{I}_4$: the mixed term cancels by ${\alpha }_i\beta =-\beta {\alpha }_i$. Therefore $P_{\pm }^{†}=P_{\pm }$ and $P_{\pm }^2=P_{\pm }$. Since $\mathrm{Tr}{\alpha }_i=\mathrm{Tr}\beta =0$, $\mathrm{Tr}{H}_{\pm }=0$ and $\mathrm{Tr}{P}_{\pm }=2$. A Hermitian projection has eigenvalues only $0$ and $1$, so its rank equals its trace and each range has dimension two. A basis can be selected for this fixed finite-dimensional range by Gaussian elimination and orthonormalized by Gram–Schmidt.

The chosen bases give ${\sum }_s{u}_s{u}_s^{†}=2E{P}_{+}=E{I}_4+H_{+}$ and ${\sum }_s{v}_s{v}_s^{†}=2E{P}_{-}=E{I}_4+H_{-}$. Multiplying on the right by $\beta$ yields $E{I}_4\beta +H_{+}\beta =E\beta -\beta \mathit{\alpha }\cdot \mathbf{p}+M{I}_4=\mathrm{p̸}+M$ and $E{I}_4\beta +H_{-}\beta =E\beta -\beta \mathit{\alpha }\cdot \mathbf{p}-M{I}_4=\mathrm{p̸}-M$, using ${\alpha }_i\beta =-\beta {\alpha }_i$ and ${\gamma }^j=\beta {\alpha }_j$.

Since $H_{+}{u}_s=E{u}_s$, multiplication by $\beta$ gives $(\mathrm{p̸}-M)u_s=0$; since $H_{-}{v}_s=E{v}_s$, it gives $(\mathrm{p̸}+M)v_s=0$. Also ${\gamma }^0({\gamma }^a{)}^{†}{\gamma }^0={\gamma }^a$: for $a=0$ this is ${\beta }^3=\beta$, and for spatial $a$ it follows from $({\gamma }^a{)}^{†}=-{\gamma }^a$ and anticommutation with $\beta$. Thus $(\bar{x}{\gamma }^{a}y{)}^{\ast }=\bar{y}{\gamma }^{a}x$ for all columns $x,y$.